2m^2=m+1 eching

$$2 m ^ { 2 } = m + 1$$

m uchun yechish

$$m=-\frac{1}{2}=-0,5$$

$$m=1$$

Guruhlash orqali faktorlash qadamlari

Ikkala tarafdan m ni ayirish.

$$2m^{2}-m=1$$

Ikkala tarafdan 1 ni ayirish.

$$2m^{2}-m-1=0$$

Tenglamani yechish uchun guruhlash orqali chap qoʻl tomonni faktorlang. Avvalo, chap qoʻl tomon 2m^{2}+am+bm-1 sifatida qayta yozilishi kerak. a va b ni topish uchun yechiladigan tizimni sozlang.

$$a+b=-1 ab=2\left(-1\right)=-2$$

ab manfiy boʻlganda, a va b da qarama-qarshi belgilar bor. a+b manfiy boʻlganda, manfiy sonda musbatga nisbatdan kattaroq mutlaq qiymat bor. Faqat bundan juftlik tizim yechimidir.

$$a=-2 b=1$$

2m^{2}-m-1 ni \left(2m^{2}-2m\right)+\left(m-1\right) sifatida qaytadan yozish.

$$\left(2m^{2}-2m\right)+\left(m-1\right)$$

2m^{2}-2m ichida 2m ni ajrating.

$$2m\left(m-1\right)+m-1$$

Distributiv funktsiyasidan foydalangan holda m-1 umumiy terminini chiqaring.

$$\left(m-1\right)\left(2m+1\right)$$

Tenglamani yechish uchun m-1=0 va 2m+1=0 ni yeching.

$$m=1 m=-\frac{1}{2}$$

Kvadrat tenglamalaridan foydalanish qoidalari

Ikkala tarafdan m ni ayirish.

$$2m^{2}-m=1$$

Ikkala tarafdan 1 ni ayirish.

$$2m^{2}-m-1=0$$

Ushbu tenglama standart shaklidadir: ax^{2}+bx+c=0. Kvadrat tenglama formulasida, \frac{-b±\sqrt{b^{2}-4ac}}{2a} 2 ni a, -1 ni b va -1 ni c bilan almashtiring.

$$m=\frac{-\left(-1\right)±\sqrt{1-4\times 2\left(-1\right)}}{2\times 2}$$

-4 ni 2 marotabaga ko'paytirish.

$$m=\frac{-\left(-1\right)±\sqrt{1-8\left(-1\right)}}{2\times 2}$$

-8 ni -1 marotabaga ko'paytirish.

$$m=\frac{-\left(-1\right)±\sqrt{1+8}}{2\times 2}$$

1 ni 8 ga qo'shish.

$$m=\frac{-\left(-1\right)±\sqrt{9}}{2\times 2}$$

9 ning kvadrat ildizini chiqarish.

$$m=\frac{-\left(-1\right)±3}{2\times 2}$$

-1 ning teskarisi 1 ga teng.

$$m=\frac{1±3}{2\times 2}$$

2 ni 2 marotabaga ko'paytirish.

$$m=\frac{1±3}{4}$$

m=\frac{1±3}{4} tenglamasini yeching, bunda ± musbat. 1 ni 3 ga qo'shish.

$$m=\frac{4}{4}$$

4 ni 4 ga bo'lish.

$$m=1$$

m=\frac{1±3}{4} tenglamasini yeching, bunda ± manfiy. 1 dan 3 ni ayirish.

$$m=-\frac{2}{4}$$

\frac{-2}{4} ulushini 2 ni chiqarib, bekor qilish hisobiga eng past shartlarga kamaytiring.

$$m=-\frac{1}{2}$$

Tenglama yechildi.

$$m=1 m=-\frac{1}{2}$$

Kvadratni yakunlash bo'yicha qadamlar

Ikkala tarafdan m ni ayirish.

$$2m^{2}-m=1$$

Ikki tarafini 2 ga bo‘ling.

$$\frac{2m^{2}-m}{2}=\frac{1}{2}$$

2 ga bo'lish 2 ga ko'paytirishni bekor qiladi.

$$m^{2}-\frac{1}{2}m=\frac{1}{2}$$

-\frac{1}{2} ni bo‘lish, x shartining koeffitsienti, 2 ga -\frac{1}{4} olish uchun. Keyin, -\frac{1}{4} ning kvadratini tenglamaning ikkala tarafiga qo‘shing. Ushbu qadam tenglamaning chap qismini mukammal kvadrat sifatida hosil qiladi.

$$m^{2}-\frac{1}{2}m+\left(-\frac{1}{4}\right)^{2}=\frac{1}{2}+\left(-\frac{1}{4}\right)^{2}$$

Kasrning ham suratini, ham maxrajini kvadratga ko'paytirib -\frac{1}{4} kvadratini chiqarish.

$$m^{2}-\frac{1}{2}m+\frac{1}{16}=\frac{1}{2}+\frac{1}{16}$$

Umumiy maxrajni topib va hisoblovchini qo'shish orqali \frac{1}{2} ni \frac{1}{16} ga qo'shing. So'ngra agar imkoni bo'lsa kasrni eng kam shartga qisqartiring.

$$m^{2}-\frac{1}{2}m+\frac{1}{16}=\frac{9}{16}$$

m^{2}-\frac{1}{2}m+\frac{1}{16} omili. Odatda, x^{2}+bx+c mukammal kvadrat bo'lsa, u doimo \left(x+\frac{b}{2}\right)^{2} omil sifatida bo'lishi mumkin.

$$\left(m-\frac{1}{4}\right)^{2}=\frac{9}{16}$$

Tenglamaning ikkala tarafining kvadrat ildizini chiqarish.

$$\sqrt{\left(m-\frac{1}{4}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Qisqartirish.

$$m-\frac{1}{4}=\frac{3}{4} m-\frac{1}{4}=-\frac{3}{4}$$

\frac{1}{4} ni tenglamaning ikkala tarafiga qo'shish.

$$m=1 m=-\frac{1}{2}$$